Answer:
G_calculated = 1.756
The outlier should be rejected, as G_cal > G_tab (= 1.463) at 95 % confidence.
Explanation:
The Grubb's test is used for identifying an outlier in data, which is from the same population. For this, a statistical term, G, is calculated for the suspected outlier. If the calculated value is greater than the tabulated G value then the suspected value is rejected. This term is given as,
G_calculated = | suspect value - mean| / s
Here, suspect value is 13.8, mean is to be taken of all the data (including suspected value). s is the standard deviation of the sample data.
s is calculated from the following formula:
s = (Σ(xi - x)²/(N-1))^1/2
Here, x is the mean, which is 15.24, xi is individual value and N is the total number of data (5).
From the above formula, s is found to be
Standard Deviation, s = 0.820
Now for G value,
G_calculated = | 13.8 - 15.24| / (0.820)
G_ calculated = 1.756
The tabulated G value at 95 % confidence and N -1 (5 - 1 = 4) degree of freedom is, 1.463.
As calculated G (1.756) is greater than the tabulated G (1.463), the value 13.8 is considered an outlier at 95 % confidence.
is formed.